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I am working my way through this book, in an attempt to teach myself matrix algebra. In the first chapter, the author asks the student to find a matrix $A$ such that:

$\begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3\end{bmatrix}\times A\times \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 4 \\ 3 & 2 & 1\end{bmatrix}$

I don't think the book went over the process for finding this solution, however.

ixtmixilix
  • 955

1 Answers1

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Let $B$ be the left matrix of the left hand side. Also, let $C$ be the right matrix of the left hand side.

Then, multiply $B^{-1}$ from the left side, and multiply $C^{-1}$ from the right side.

Edit : Since we have $$BAC=D$$ where $D$ is a matrix in the right hand side, we have $$B^{-1}BACC^{-1}=B^{-1}DC^{-1}.$$ Since each of $B^{-1}B$ and $CC^{-1}$ is an identity matrix, we have $$A=B^{-1}DC^{-1}.$$ I hope this helps.

mathlove
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